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3 years ago

Work out the length of the following rectangles

Mathematics

1 answer:

tiny-mole [99]3 years ago

8 0

(a) a = 10.9 cm

(b) b = 20.78 cm

(c) c = 4.06 cm

<u>Explanation:</u>

(a)

height, h = 5cm

Hypotenuse, H = 12cm

Base, a = ?

Using pythagoras theorm:

(H)² = (h)² + (a)²

(12)² = (5)² + (a)²

119 = a²

a = 10.9 cm

(b)

Hypotenuse, H = 24 cm

Base, a = 12 cm

Height, b = ?

Using pythagoras theorm:

(H)² = (b)² + (a)²

(24)² = (b)² + (12)²

432 = b²

b = 20.78 cm

(c)

height, h = 4 cm

Hypotenuse, H = 5.7 cm

Base, c = ?

Using pythagoras theorm:

(H)² = (h)² + (c)²

(5.7)² = (4)² + (c)²

16.49 = c²

c = 4.06 cm

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Mrrafil [7]

Answer:

$E(X_n)=\frac{2(n-1)}{27}$

$E(y)=\frac{14}{9}$

Step-by-step explanation:

From the question we are told that:

Sample size n=9

Number of Green $g=4$

Number of yellow $y=4$

Number of white $w=4$

Probability of Green Followed by yellow P(GY) ball

$P(GY)=\frac{4}{9}*\frac{3}{9}$

$P(GY)=\frac{4}{27}$

Generally the equations for when n is even is mathematically given by

$Probability of success P(S)=\frac{4}{27}$

$Probability of Failure P(F)=\frac{27-4}{27}$

$Probability of Failure P(F)=\frac{23}{27}$

Therefore

$E(X_n)=\frac{n}{2}*P$

$E(X_n)=\frac{n}{2}*\frac{4}{27}$

$E(X_n)=\frac{2n}{27}$

Generally the equations for when n is odd is mathematically given by

$\frac{n-1}{2}$

$E(X_n)=\frac{n-1}{2}*\frac{4}{27}$

$E(X_n)=\frac{2(n-1)}{27}$

Probability of drawing white ball

$P(w)=\frac{2}{9}$

Therefore

$E(w)=\frac{1}{p}$

$E(w)=\frac{1}{\frac{2}{9}}$

$E(w)=\frac{9}{2}$

Therefore

$E(y)=[E(w)-1]\frac{4}{9}$

$E(y)=[\frac{9}{2}-1]\frac{4}{9}$

$E(y)=\frac{14}{9}$

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now, let's multiply everything by 2:

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so the answer is -32!

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